explanation added for 21 DP

Author: arorarahul

README.md

@@ -385,6 +385,7 @@ it down into a story and later generalize to form recursive equations
 - [Find the number of n-bit integers which do not have any two consequent zeroes](/dynamic-programming/question18.c)
 - [Given a sentence without spaces between words. Break the words in such a way that they are meaningful](/dynamic-programming/question19.c)
 - [Partition problem](/dynamic-programming/question20.c)
+- [Find the longest palindromic subsequence](/dynamic-programming/question21.c)
 
 ## Some important concepts to solve algos better
 

dynamic-programming/question21.c

@@ -0,0 +1,48 @@
+/*
+Find the longest palindromic subsequence
+
+Naive approach:
+Find all the subsequences of a given string. 2^n.
+Then O(n) time to see if its a palindrome or not and then keep a variable to find the max
+length. This time complexity is exponential
+
+METHOD1:
+DP: We can reverse the string and store it in an another string.
+Now we can apply the algo to find the LCS. Whatever the LCS will be is going to be a palindrome 
+itself.
+
+Time complexity: O(n^2)
+
+METHOD:
+DP: Explanation:
+Here we can consider a string and try to match the first and the last element, like we do in order
+to identify whether a string is a palindrome or not.
+We will have two possibilities
+
+if it matches    (then we move the pointer ahead to compare the inner elements from both sides)
+if it does not match (then we take max of two cases. In one we move pointer from right and keep left fixed
+and other is opposite of it)
+
+Recursive equation for the same can be formed as below:
+
+PL(0,n-1) = {
+		
+		PL(1,n-2) //if equal
+		max{ PL(0,n-2) , PL(1,n-1)  } //if not equal
+}
+
+Now it is proved that it can be solved recursively. Now taking an example for 0,5 and making a tree
+using these recursive equations will show us repeating sub problems.
+
+Therefore, we need to find unique sub problems, for that we generalize the definition of recursive eqns
+as:
+PL   (i,   j)
+if    1    n-1  //if value of i is 1 j can be between 1 and n-1, therefore it can have n-1 values
+      2    n-2
+      3	   n-3 and so on 
+
+      That means there are n^2 unique sub problems
+
+Therefore, we make a matrix of size n^2 and ...
+
+*/